lect 3 by dr irshad ahmad civial dep.pptx

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1. 1 CE-313 (2 Credit Hours) Geotechnical Engineering-II Ch-2: Consolidation Instructor: Prof- Dr Irshad Ahmad Lecture-3 Department of Civil Engineering University of Engineering and Technology, Peshawar

2. 2 CONTENTS Consolidation settlement calculations for NCC and OCC Coefficient of volume compressibility

3. 3 Soil Solids Voids V o V v V s ∆V Prove that: ∆H/H o = (e o -e 1 )/(1+e o ) Soil Solids Voids H o H s ∆H H v Soil Solids Voids e o 1+e o ∆e 1 V s , H s = Volume, Height of soil solids V v , H v = Volume, height of soil voids ∆V, ∆H = change in Volume, height e o = initial void ratio = V v /V s If cross sectional area of the soil sample=1 (1-D consolidation) e o = H v /H s If Hs is assumed to be 1 then e o = H v ∆H/H o = ∆e/(1+eo) = (e o -e 1 )/(1+e o )

4. 4 EXAMPLE eo=1 e1=0.8 Ho=10m Fill Load o  Hc

5.   5 Coefficient of Volume Compressibility ( m v )

6. The value of m v for a particular soil is not constant but depends on the stress range over which it is calculate. BS 1377 specifies the use of the coefficient m v calculated for a stress increment of 100 kN/m 2 in excess of the effective overburden pressure of the in-situ soil at the depth of interest, although the coefficient may also be calculated, if required, for any other stress range. 6 Coefficient of Volume Compressibility ( m v ) e o e 1 e ꞌ o ꞌ 1 ꞌ 100kPa

7. 7 1D Oedometer Test Data Analysis

8. A consolidation test was conducted on a saturated clay ( G s =2.73). The initial thickness of the specimen was 19.0mm and at the end of the test the specimen was removed from the apparatus and the water content was determined to be 19.8%. If during the whole test the change in thickness of the specimen was 3.52mm. Find the change in void ratio. 8 EXAMPLE Given Data: Gs = 2.73 w 1 = natural moisture content at the end of test = 19.8% S=1 (100% saturation assumed) H= Total change in sample height= 3.52mm Void Ratio at the end of test = e 1 *S = G s *w 1 e 1 = (w 1 G s )/S = (0.198 x 2.73)/1 = 0.541 Void ratio at the start of the test = e o = e 1 + ∆e Now 3.52mm 19mm 0.541 ∆e = 0.350 e o = e 1 + ∆e = 0.541 + 0.35 = 0.891 (∆e/∆H)=[(1+0.891)/19] (∆e/∆H)=1.891/19 ∆e = (1.891/19)∆H

9. The following compression readings were obtained in an oedometer test on a specimen of saturated clay (G s =2.73); The initial thickness of the specimen was 19.0mm and at the end of the test the water content was 19.8%. Plot the e-log ꞌ curve and Determine the preconsolidation pressure. Determine the values of m v for the stress increments 100-200 and 1000-1500 kPa. What is the value of C c for the latter increment? 9 Pressure (kN/m 2 ) 54 107 214 429 858 1716 3432 Dial Gauge after 24 hrs.(mm) 5.00 4.747 4.493 4.108 3.449 2.608 1.676 0.737 1.480 EXAMPLE

10. 10 Solution e-log( ꞌ) graph Given Data: Gs = 2.73 w 1 = natural moisture content at the end of test = 19.8% S=1 (100% saturation assumed) H= Total change in sample height= 5-1.48=3.52mm Void Ratio at the end of test = e 1 *S = G s *w e 1 = (w 1 G s )/S = (0.198 x 2.73)/1 = 0.541 Void ratio at the start of the test = e o = e 1 + ∆e Now e o = e 1 + ∆e = 0.541 + 0.35 = 0.891 (∆e/∆H)=[(1+0.891)/19] (∆e/∆H)=1.891/19 ∆e = (1.891/19)∆H 3.52mm 19mm 0.541 ∆e = 0.350

11. 11 Pressure(kN/m 2 ) Dial Gauge after 24 hrs.(mm) H ∆H(mm) ∆e = (1.891/19)* ∆H e = e o -∆e e = 0.891 - ∆e 5.00 0.891 54 4.747 5 - 4.747 = 0.253 0.025 0.866 107 4.493 5 - 4.493 = 0.507 0.05 0.841 214 4.108 5 - 4.108 = 0.892 0.089 0.802 429 3.449 5 - 3.449 = 1.55 0.154 0.737 858 2.608 5 – 2.608 = 2.392 0.238 0.653 1716 1.676 5 – 1.676 = 3.324 0.331 0.560 3432 0.737 5 – 0.737 = 4.263 0.424 0.467 1.480 5 – 1.480 = 3.52 0.350 0.541 Solution ∆e = (1.891/19)∆H

12. 12 (a) e-log( ) curve

13. 13 Point of maximum curvature Horizontal Bisector Tangent Prdoced back straight line ꞌ p =325 kPa (b) Pre-consolidation pressure  p 

14. From e-logꞌ graph the void ratios at the given loads are; e o = 0.845 & e 1 = 0.808 respectively Therefore ∆e = e o -e 1 =0.845-0.808=0.037 Put this value in m v m v = [(e o -e 1 )/(1+e o )]/(ꞌ 1 - ꞌ o ) m v = [0.037/(1+0.845)]/(200-100) m v = 2.0x10 -4 m 2 /MN 14 m v for stress increment of 100-200 kPa e o =0.845 ꞌ o =100 kPa e 1 =0.808 ꞌ 1 =200 kPa

15. 15 m v for stress increment of 1000-1500 kPa From e-logꞌ graph the void ratios at the given loads are; e o = 0.632 & e 1 = 0.577 respectively Therefore ∆e = e o -e 1 =0.632-0.577=0.055 Put this value in m v m v = [(e o -e 1 )/(1+e o )]/(ꞌ 1 - ꞌ o ) m v = [0.055/(1+0.632)]/(1500-1000) m v = 0.067x10 -4 m 2 /MN e o =0.632 ꞌ o =1000 kPa e 1 =0.577 ꞌ 1 =1500 kPa

16. 16 Value of C c for the Later increment (1000-1500 kPa ) ꞌ =1000 kPa ꞌ 1 =1500kPa ∆e = e o -e 1 =0.632-0.577=0.055 C c = [∆e/log( ꞌ 1 / ꞌ )] C c = [0.055/log(1500/1000)] C c = 0.31 e o =0.632 ꞌ o =1000 kPa e 1 =0.577 ꞌ 1 =1500 kPa Note: The value of C c will be same for any stress range on the linear part of the e-log ꞌ curve while m v will vary according to the stress ranges, even on the linear part of the curve.

17. 17 Consolidation Settlement (∆Hc) for NCC ? For NCC the compression index “Cc” is Cc = [∆e/log(( ꞌ o + ∆ꞌ)/ꞌ o )] ∆e = Cc * log[( ꞌ o + ∆ꞌ)/ꞌ o ] H o = height of the consolidating layer ꞌ o = effective overburden pressure ∆= change in total stress e o = Initial void ratio NCC eo, Cc Ho ∆ ꞌ = ꞌ 1 - ꞌ o  Hc ꞌ o ꞌ 1 e o ∆ logꞌ e 1

18. 18 Consolidation Settlement-OCC

19. 19 Consolidation Settlement-OCC

20. 20 Estimate the consolidation settlement (∆H c ) for a10m layer of NCC. Initial void ratio is 2.5, preconsolidation pressure is 7kPa, the average stress increase due to application of the footing is 10kPa. Given Data Clay layer = 10m eo = 2.5 ꞌ p = 7kPa ∆ꞌ = 10kPa C c = 0.986 Required Consolidation Settlement ∆H c = ? Calculation Since the soil is normally consolidated clay, therefore ꞌ o = ꞌ p =7kPa put given data into the below equation ∆H c = C c [H o /(1+e o )]*Log[(ꞌ o +∆ꞌ)/ꞌ o ] ∆H c = 0.986[10/(1+2.5)]*Log[(7+10)/7] ∆H c = 1.09 m EXAMPLE Clay Cc=0.986 e o =2.5 10 m ∆ ꞌ = 10 kPa ꞌ o =7 kPa 7+10=17kPa e o =2.5 ∆ =10 kPa logꞌ

21. 21 EXAMPLE 4-3 Silty Clay 10 m ∆ ꞌ = 35 kPa e o =0.84 ꞌ o =80kPa ꞌ p =130 kPa ∆ =35kPa 115 kPa logꞌ Find ∆Hc for a 10 m thick silty clay layer. ꞌ o = 80 kPa, ꞌ p = 130 kPa, the structural loads at the surface will increase the average stress in the clay layer by 35 kPa. C r = 0.03, e o = 0.84 Solution Case-1 ꞌ o + ∆ꞌ =80+35=115 kPa < ꞌ p

22. 22 EXAMPLE 4-4 Silty Clay 10 m ∆ ꞌ = 90 kPa Same data as in example 4-3 but now the building loads produce an average stress increase of 90 kPa. C c = 0.25; C r = 0.03 Solution ꞌ o +∆ ꞌ =80 + 90 =170 kPa > ꞌ pc Put given values in above equation ∆H c = 0.034 + 0.158 =0.193 m e o =0.84 ꞌ o =80kPa ꞌ p =130 kPa ∆ =90kPa 170 kPa logꞌ

23. Step-1 Calculate mv from laboratory test using Figure below. Step-2 Use equation Hc = mv Ho  23 Ho mv  How to calculate  Hc from mv e o e 1 e ꞌ o ꞌ 1 ꞌ

24. Fill load acting over very large area produces pressure of 125kPa. The soil profile is given in diagram. The value of m v for the clay is 0.35 m 2 /MN. Determine the final consolidation settlement due to the surface load. Solution As the load extends over very large area so consolidation in clay layer can be assumed to be 1-dimensional. Since the consolidation settlement is calculated in terms of mv only the increase in effective stress due to fill load at mid depth of clay layer is required (the increment being assumed constant over the depth of the clay layer and equal to fill load) 24 Sand Clay mv=0.35 m 2 /MN 10m 4m 125 kPa EXAMPLE 4-5 As we know ∆H c = m v *∆ ꞌ *H o ∆H c = 0.35*125*4 =175mm